Three vectors P→,Q→ and R→ are shown in the figure. Let S be any point on the vector R→ . The distance between the points P and S is bR→. The general relation among vectors P→,Q→ and S→ and is

S¯+(1−b)(Q¯−P¯)=Q¯S¯=Q¯(1−1+b)+(1−b)P¯=(1−b)P¯+bQ¯

Three vectors P→,Q→ and R→ are shown in the figure. Let S be any point on the vector R→ . The distance between the points P and S is bR→. The general relation among vectors P→,Q→ and S→ and is

S¯+(1−b)(Q¯−P¯)=Q¯S¯=Q¯(1−1+b)+(1−b)P¯=(1−b)P¯+bQ¯

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Three vectors $\vec{P}, \vec{Q} and \vec{R}$ are shown in the figure. Let S be any point on the vector $\vec{R}$ . The distance between the points P and S is $b |\vec{R}|$ . The general relation among vectors $\vec{P}, \vec{Q} and \vec{S}$ and is

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$\vec{S} = (1 - b) \vec{P} + b \vec{Q}$

$\vec{S} = (1 - b) \vec{P} + b^{2} \vec{Q}$

$\vec{S} = (1 - b^{2}) \vec{P} + b \vec{Q}$

$\vec{S} = (b - 1) \vec{P} + b \vec{Q}$

answer is C.

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$\begin{array}{l} \bar{S} + (1 - b) (\bar{Q} - \bar{P}) = \bar{Q} \\ \bar{S} = \bar{Q} (1 - 1 + b) + (1 - b) \bar{P} \\ = (1 - b) \bar{P} + b \bar{Q} \end{array}$

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Three vectors P→,Q→ and R→ are shown in the figure. Let S be any point on the vector R→ . The distance between the points P and S is bR→. The general relation among vectors P→,Q→ and S→ and is

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Three vectors $\vec{P}, \vec{Q} and \vec{R}$ are shown in the figure. Let S be any point on the vector $\vec{R}$ . The distance between the points P and S is $b |\vec{R}|$ . The general relation among vectors $\vec{P}, \vec{Q} and \vec{S}$ and is